42x^2-48x+12=0

Solution for 42x^2-48x+12=0 equation:

42x^2-48x+12=0

a = 42; b = -48; c = +12;
Δ = b2-4ac
Δ = -482-4·42·12
Δ = 288
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-12\sqrt{2}}{2*42}=\frac{48-12\sqrt{2}}{84}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+12\sqrt{2}}{2*42}=\frac{48+12\sqrt{2}}{84}
$